At the top of a cliff 80 m high,Raoul throws a rock upward with velocity 11.0 m/s.How much later should he drop a second rock from rest so both rocks arrive simultaneously at the bottom of the cliff?

Question

At the top of a cliff 80 m high,Raoul throws a rock upward with velocity 11.0 m/s.How much later should he drop a second rock from rest so both rocks arrive simultaneously at the bottom of the cliff?
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Quizplus · Accepted Answer

The first rock's motion can be described by the equation $ h = v_0t + \frac{1}{2}gt^2 $, where $ h = -80 $ m, $ v_0 = 11 $ m/s, and $ g = -9.8 $ m/s². Solving for $ t $ gives approximately 4.34 s. The second rock is dropped from rest, so its time to fall is $ t_2 = \sqrt{\frac{2h}{g}} \approx 4.34 $ s. The difference in time is 4.34 s - 3.07 s = 1.27 s.

At the Top of a Cliff 80 M High,Raoul Throws